Substitute $x=1$ into $(1+x)^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}x^k$ to obtain $2^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}=2\sum_{k=0}^{n}\binom{2n+1}{k}$ since $\binom{2n+1}{k}=\binom{2n+1}{2n+1-k}$.